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a(n) = 13*a(n-1) - 65*a(n-2) + 156*a(n-3) - 182*a(n-4) + 91*a(n-5) - 13*a(n-6).
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%I #70 Feb 15 2024 08:45:32

%S 0,-1,-5,-22,-91,-364,-1430,-5564,-21541,-83200,-321100,-1239446,

%T -4787770,-18514119,-71683040,-277913233,-1078918139,-4194134516,

%U -16324764560,-63616690111,-248187382924,-969250588865,-3788814577730,-14823325196459,-58040165033110,-227415509487686

%N a(n) = 13*a(n-1) - 65*a(n-2) + 156*a(n-3) - 182*a(n-4) + 91*a(n-5) - 13*a(n-6).

%C a(n) is equal to the rational part of 2*X(2*n)/sqrt(13) (with respect of the field Q(sqrt(13))), where X(n) = sqrt((13 + 3*sqrt(13))/2)*X(n-1) - sqrt(13)*X(n-2) + sqrt((13 - 3*sqrt(13))/2)*X(n-3), with X(0)=3, X(1)=sqrt((13 + 3*sqrt(13))/2), and X(2)=(13 - sqrt(13))/2.

%C The Berndt-type sequence number 4 for the argument 2Pi/13 defined by the relation A216508(n) + a(n)*sqrt(13) = 2*X(2*n), where X(n) := s(2)^n + s(5)^n + s(6)^n, where s(j) := 2*sin(2*Pi*j/13).

%C I observe that all numbers of the form (a(6*n + k + 4) - 4*a(6*n + k + 3))*13^(-n), where k = 1,...,6, n = 0,1,... are integers. For example we have a(10)-4*a(9)=900*13 and a(11)-4*a(10)=266*13^2.

%C We note that a(n) = -A050185(n) for n=0,1,...,5 and a(6) + A050185(6) = -2. - _Roman Witula_, Sep 22 2012

%C a(n) is equal to the negative rational part of 2*Y(2*n)/sqrt(13) (with respect of the field Q(sqrt(13))), where Y(n) = sqrt((13 - 3*sqrt(13))/2)*Y(n-1) + sqrt(13)*Y(n-2) - sqrt((13 + 3*sqrt(13))/2)*Y(n-3), with Y(0)=3, Y(1)=sqrt((13 - 3*sqrt(13))/2), and Y(2)=(13 + sqrt(13))/2. It can be proved that Y(n) = s(1)^n + s(3)^n + s(9)^n (we have s(9) = -s(4)), and 2*Y(2*n) = A216508(n) - a(n)*sqrt(13). - Roman Witula, Sep 24 2012

%D R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, Thirteenth International Conference on Fibonacci Numbers and their Applications, Congressus Numerantium, 201 (2010), 89-107.

%D R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

%H Andrew Howroyd, <a href="/A216597/b216597.txt">Table of n, a(n) for n = 0..200</a>

%H G. Dresden and Y. Li, <a href="https://arxiv.org/abs/2210.04322">Periodic Weighted Sums of Binomial Coefficients</a>, arXiv:2210.04322 [math.NT], 2022.

%H R. Witula and D. Slota, <a href="https://www.mathstat.dal.ca/fibonacci/abstracts.pdf">Quasi-Fibonacci numbers of order 13</a>, (abstract) see p. 15.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (13,-65,156,-182,91,-13).

%F G.f.: -x*(13*x^4 - 26*x^3 + 22*x^2 - 8*x + 1) / (13*x^6 - 91*x^5 + 182*x^4 - 156*x^3 + 65*x^2 - 13*x + 1). - _Colin Barker_, Jun 01 2013

%F a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,n+k)*(k|13), where (k|13) represents the Legendre symbol. - _Greg Dresden_, Oct 09 2022

%e We have s(2)^4 + s(5)^4 + s(6)^4 + sqrt(13) = s(2)^2 + s(5)^2 + s(6)^2 = (13 - sqrt(13))/2.

%e We note that 2*a(1) - a(2) = 1, 4*a(2) - a(3) = 2, 4*a(3) - a(4) = 3, 4*a(4) = a(5) and 4*a(n) - a(n+1) < 0 for every n = 5,6,...

%t LinearRecurrence[{13,-65,156,-182,91,-13}, {0,-1,-5,-22,-91,-364}, 30]

%o (PARI) concat([0], Vec(-x*(13*x^4 -26*x^3 +22*x^2 -8*x +1) / (13*x^6 -91*x^5 +182*x^4 -156*x^3 +65*x^2 -13*x +1) + O(x^30))) \\ _Andrew Howroyd_, Feb 25 2018

%Y Cf. A216605, A216486, A216508, A216540, A161905, A216801.

%K sign,easy

%O 0,3

%A _Roman Witula_, Sep 11 2012

%E Better name from _Joerg Arndt_, Sep 17 2012